The electric field intensity on the surface of a charged sphere with a radius R and charge of 3×10^-6 C, where the electric potential is 500V, can be calculated using the formula E = (V^2)/(q/(4πε0)). The correct expression for electric potential V is V = 1/(4πε0) * (q/R), not squared as initially stated. The derived formula for electric field intensity E simplifies to E = (1/(4πε0)) * (q/R^2), confirming the relationship between electric potential and electric field strength.
PREREQUISITESStudents studying electromagnetism, physics educators, and anyone interested in understanding the principles of electric fields and potentials in charged objects.
##F##=##\frac{K Qq}{r^2}##ehild said:Do you remember Coulomb's Law?
They took the equation ##E##=##\frac{1}{4πε_0}\frac{q}{R^2}## and rewrote it in that form. You can check they are equivalent. This was useful because the numerator is the same as V2 and the denominator only involves the charge and some constants. Since V and q are known, this gives a quick way of finding E.gracy said:I am still clueless about my question in op.
##E##=##\frac{\left(\frac{1}{4πε0}\frac{q}{R}\right)^2}{\frac{q}{4πε0}}##